In the figure (ii) given below, ABC is an isosceles triangle in which AB = AC and Q is mid-point of AC. If APB is a secant and AC is tangent to the circle at Q, prove that AB = 4AP.

We know that,

If a chord and a tangent intersect externally, then the product of lengths of the segments of the chord is equal to the square of the length of the tangent from the point of contact to the point of intersection.

∴ AQ² = AP × AB …..(Eq. 1)

Given Q is mid-point of AC, so AQ = $\dfrac{AC}{2}$.

$\Rightarrow \Big(\dfrac{AC}{2}\Big)^2 = AP \times AB$

Since, AC = AB.

$\Rightarrow \Big(\dfrac{AB}{2}\Big)^2 = AP \times AB \\[1em] \Rightarrow \dfrac{AB^2}{4} = AP \times AB \\[1em] \Rightarrow AB^2 = 4 \times AP \times AB$

Dividing both sides by AB,

AB = 4 x AP.

Hence, proved that AB = 4AP.